\subsection{答案}
\subsubsection{例题}
\begin{enumerate}
    \item 方法一：\[S(D_1)=2\int_0^2\sqrt{2x}\mathrm{d}x=2\sqrt2\int_0^2\sqrt x\mathrm{d}x=\frac{16}3,\]
    \[S(D_2)=\int_2^8(\sqrt{2x}-(x-4))\mathrm{d}x=\frac{38}3.\]
    累加，可以得到：
    \[S(D)=S(D_1)+S(D_2)=18.\]
    方法二：\[S(D)=\int_{-2}^4\left(y+4-\frac12y^2\right)\mathrm{d}y=18.\]
    \item \[\begin{aligned}S&=\left.2\cdot\frac12\int_0^\pi a^2\left(1+\cos\theta\right)^2\mathsf{d}\theta\right.\\&=a^2\int_0^\pi(1+2\mathsf{cos~}\theta+\cos^2\theta)\mathsf{d}\theta\\&=\frac32\pi a^2.\end{aligned} \]
    \item 因为\[
    \begin{aligned}&x^{\prime}(t)=-3a\cos^2t\sin t,\\&y^{\prime}(t)=3a\sin^2t\cos t,\end{aligned}
    \]，根据弧长公式，得
    \[
    \begin{aligned}S(\Gamma)&=4\int_0^{\pi/2}\sqrt{(x^{\prime}(t))^2+(y^{\prime}(t))^2}\mathrm{d}t\\&=12a\int_0^{\pi/2}\cos t\sin t\mathrm{d}t=6a.\end{aligned}
    \]
    \item \[\begin{aligned}S&=\int_0^{2\pi}\sqrt{a^2\left(1+\cos\theta\right)^2+a^2\sin^2\theta}\mathrm{d}\theta\\&=\left.2a\int_0^{2\pi}\left|\cos\frac\theta2\right|\mathrm{d}\theta\right.=\left.4a\int_0^{\pi}\cos\frac\theta2\mathrm{d}\theta\right.=\left.8a\right..\end{aligned}
    \]
    \item 记两个柱面所交的部分为 $\Omega$.由对称性知，Ω 的体积是它在第一卦限中那部分体积的 8倍.因此我们只需计算它在第一卦限部分的体积.容易知道，当$0\leq x\leq a$时，以横坐标为$x$的平面截第一卦限中那部分所得的截面是以$\sqrt{a^2-x^2}$为边的正方形,其面积为$$g(x)=a^2-x^2.$$，所以为体积$V(\Omega) = 8\int _0^a( a^2- x^2) $d$x= \frac 163a^3.$
    \item 把球体看作\(\begin{aligned}
        y=\sqrt{a^2-x^2}(-a\leq x\leq a)
    \end{aligned}\)围绕$x$旋转得到，所以
    \[V=\left.\pi\int_{-a}^a(a^2-x^2)\mathrm{d}x\right.=\left.2\pi\right]_0^a(a^2-x^2)\mathrm{d}x=\frac43\pi a^3\]
    \item 把球面看作\(\begin{aligned}
        y=\sqrt{a^2-x^2}(-a\leq x\leq a)
    \end{aligned}\)围绕$x$旋转得到，所以
    \[S=2\pi\int_{-a}^a\sqrt{a^2-x^2}\sqrt{1+\frac{x^2}{a^2-x^2}}\mathrm{d}x=2\pi a\int_{-a}^a\mathrm{d}x=4\pi a^2\]
    \item \begin{enumerate}[(1)]
        \item 已经算得结果了
        \item 心脏线得参数方程可以写成\[
        \begin{cases}x=a(1+\cos\theta)\cos\theta\\[1ex]y=a(1+\cos\theta)\sin\theta\end{cases}
        \],所以体积为\(\begin{aligned}
            \pi\int_{0}^{2a}y^{2}\mathrm{d}x=\pi a^{3}\int_{\pi}^{0}((1+\cos\theta)\sin\theta)^{2}\mathrm{d}(1+\cos\theta)\cos\theta=\frac{8\pi a^{3}}{3}.
        \end{aligned}\)
    \end{enumerate}
    \item \begin{enumerate}[(1)]
        \item \[
        \int_0^{+\infty}\mathrm{e}^{-ax}\cos\left.bx\mathrm{d}x\right.=\frac a{a^2+b^2}
        \]
        \item \[\int_0^{+\infty}\mathrm{e}^{-ax}\sin^2bx\mathrm{d}x^2=\frac b{a^2+b^2}\]
    \end{enumerate}
    \item 答案是\(\begin{aligned}
        \frac{\pi}{\sqrt{2}}
    \end{aligned}\).\textbf{扩展}：计算\(\begin{aligned}
        \int \frac{1}{x^4+1}\mathrm{d}x
    \end{aligned}\)的定积分：
    \[
    \begin{aligned}
        \begin{aligned}\int\frac{\mathrm{d}x}{1+x^4}&=\frac{1}{2}\int\frac{1+x^2}{1+x^4}\mathrm{d}x+\frac{1}{2}\int\frac{1-x^2}{1+x^4}\mathrm{d}x\\&=\begin{aligned}\frac{1}{2}\int\frac{\mathrm{d}(x-1/x)}{(x-1/x)^2+2}-\frac{1}{2}\int\frac{\mathrm{d}(x+1/x)}{(x+1/x)^2-2}\end{aligned}\\&=\frac{1}{2\sqrt{2}}\arctan\frac{x-1/x}{\sqrt{2}}-\frac{1}{4\sqrt{2}}\ln\left|\frac{x^2-\sqrt{2}x+1}{x^2+\sqrt{2}x+1}\right|+C\end{aligned}
    \end{aligned}
    \]
    \item 这里 $a$ 与$b$ 都是瑕点.容易看出，这个瑕积分是收敛的.我们用换元法来计算它的值.当$x\in(a,b)$时，$$\frac{x-a}{b-a},\quad\frac{b-x}{b-a}
$$是正数，并且其和等于 1,因此可设$$\frac{x-a}{b-a}=\sin^2\theta\quad\left(0<\theta<\frac\pi2\right).$$
此时，
$x=a+(b-a)\sin^2\theta=a\cos^2\theta+b\sin^2\theta$, d$x= 2( b- a) \cos \theta\sin \theta$d$\theta.$于是，瑕积分可化为常义下的积分：$$2\int_0^{\pi/2}\mathbf{d}\theta=\pi$$
\end{enumerate}